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1. \(f\) is injective.

   Proof.

   Let \(x\) and \(y\) be elements of the domain \(\Z\text{.}\) Assume \(f(x) = f(y)\text{.}\) If \(x\) and \(y\) are both even, then \(f(x) = x+1\) and \(f(y) = y+1\text{.}\) Since \(f(x) = f(y)\text{,}\) we have \(x + 1 = y + 1\) which implies that \(x = y\text{.}\) Similarly, if \(x\) and \(y\) are both odd, then \(x - 3 = y-3\) so again \(x = y\text{.}\) The only other possibility is that \(x\) is even an \(y\) is odd (or visa-versa). But then \(x + 1\) would be odd and \(y - 3\) would be even, so it cannot be that \(f(x) = f(y)\text{.}\) Therefore if \(f(x) = f(y)\) we then have \(x = y\text{,}\) which proves that \(f\) is injective.

2. \(f\) is surjective.

   Proof.

   Let \(y\) be an element of the codomain \(\Z\text{.}\) We will show there is an element \(n\) of the domain (\(\Z\)) such that \(f(n) = y\text{.}\) There are two cases: First, if \(y\) is even, then let \(n = y+3\text{.}\) Since \(y\) is even, \(n\) is odd, so \(f(n) = n-3 = y+3-3 = y\) as desired. Second, if \(y\) is odd, then let \(n = y-1\text{.}\) Since \(y\) is odd, \(n\) is even, so \(f(n) = n+1 = y-1+1 = y\) as needed. Therefore \(f\) is surjective.